We have to find the value of the expression 
We know that the below values.
Hence, in order to find the value of the given expression, we can first rewrite it in terms of 
Now, we know that 
Hence, we have
C is the correct option.
The circumference of the top of the cake is 
in.
It sounds like the baker is trying to line the circumference with cherriesto fit as many as possible along the border, like in the picture I've attached below.
This comes down to dividing the cake's circumference by the diameter of each cherry:
or about 39 cherries, making the answer D.
Answer:
There is not sufficient evidence to support the claim.
Step-by-step explanation:
The claim to be tested is:
The mean respiration rate (in breaths per minute) of students in a large statistics class is less than 32.
To test this claim the hypothesis can be defined as follows:
<em>H₀</em>: The mean respiration rate of students is 32, i.e. <em>μ</em> = 32.
<em>Hₐ</em>: The mean respiration rate of students is less than 32, i.e. <em>μ</em> < 32.
The sample mean respiration rate of students is 31.3.
According to the claim the sample mean is less than 32.
The sample mean value is not unusual if the claim is true, and the sample mean value is also not unusual if the claim is false.
Thus, there is not sufficient evidence to support the claim.
Answer:
x = 44
Step-by-step explanation:
sqrt(x+5) -3 =4
Add 3 to each side
sqrt(x+5) -3+3 =4+3
sqrt(x+5) =7
Square each side
(sqrt(x+5))^2 =7^2
x+5 = 49
Subtract 5 from each side
x+5-5 = 49-5
x=44
Answer:
(5.4582 ; 6.8618)
Step-by-step explanation:
Given the data:
6 10 2 6 3 3 3 6 6 6 6 5 8 9 10 10 7 9 3 6 5 10 9 9 10 3 8 6 6 3 3 6 6 5 4 10 9 3 5 7 10 6 3 8 6 8 3 3 5 5
Sample mean, xbar = Σx / n
n = sample size = 50
ΣX = 308
xbar = 308 / 50 = 6.16
Using a Calculator :
The sample standard deviation, s = 2.469
Confidence interval = xbar ± margin of error
Margin of Error = Tcritical * s/sqrt(n)
Tcritical at 95% ; df = 50 - 1 = 49
Tcritical = 2.010
Hence,
Margin of Error= 2.010 * (2.469/sqrt(50)) = 0.7018
Lower boundary : (6.16 - 0.7018) = 5.4582
Upper boundary : (6.16 + 0.7018) = 6.8618
(5.4582 ; 6.8618)
